1977 Canadian MO Problems/Problem 1
Revision as of 21:07, 25 July 2006 by Boy Soprano II (talk | contribs)
Problem
If prove that the equation has no solutions in positive integers and
Solution
Directly plugging and into the function, We now have a quadratic in
Applying the quadratic formula,
In order for both and to be integers, the discriminant must be a perfect square. However, since the quantity cannot be a perfect square when is an integer. Hence, when is a positive integer, cannot be.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.